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3 years ago

4 Compare and contrast the reactants and products in the two combustion

Chemistry

1 answer:

netineya [11]3 years ago

7 0

Answer:

methane + oxygen → carbon dioxide + water

Explanation:

methane + oxygen → carbon dioxide + water methane + oxygen → carbon dioxide + water methane + oxygen → carbon dioxide + water

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2,6 -dimetil nonano<br> 4 etil-2,2dimetil-4-propo:0ctano

Svetllana [295]

Answer:

it will be 26 %t _4

Explanation:

do the math and tst

7 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

What is the least polar solvent /solvent system used to run the tlc?

Lynna [10]

I believe that it is petroleum ether.

8 0

3 years ago

Plz I need help on this Science question I don’t understand

sashaice [31]

Answer:

Explanation:

isjshsjsussibs8sjbs

7 0

3 years ago

22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass

Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0

3 years ago